Overview
Fundamental principle of counting
Multiplication principle (Fundamental Principle of Counting)
Suppose an event \(E\) can occur in \(m\) different ways and associated with each way of occurring of \(E\), another event \(F\) can occur in \(n\) different ways, then the total number of occurrence of the two events in the given order is \(m \times n\).
Addition principle
If an event \(E\) can occur in \(m\) ways and another event \(F\) can occur in \(n\) ways, and suppose that both can not occur together, then \(E\) or \(F\) can occur in \(m+n\) ways.
Permutations
A permutation is an arrangement of objects in a definite order.
Permutation of \(n\) different objects
The number of permutations of \(n\) objects taken all at a time, denoted by the symbol \({ }^n P _n\), is given by
\(
{ }^n P _n=n!
\)
where \(n!=n(n-1)(n-2) \ldots 3.2 .1\), read as factorial \(n\), or \(n\) factorial.
The number of permutations of \(n\) objects taken \(r\) at a time, where \(0<r \leq n\), denoted by \({ }^n P _r\), is given by
\(
{ }^n P _r=\frac{n!}{(n-r)!}
\)
\(
\text { We assume that } 0!=1
\)
When repetition of objects is allowed
The number of permutations of \(n\) things taken all at a time, when repetition of objects is allowed is \(n^n\).
The number of permutations of \(n\) objects, taken \(r\) at a time, when repetition of objects is allowed, is \(n^r\).
Permutations when the objects are not distinct
The number of permutations of \(n\) objects of which \(p_1\) are of one kind, \(p_2\) are of second kind, …, \(p_k\) are of \(k^{\text {th }}\) kind and the rest if any, are of different kinds is \(\frac{n!}{p_{1}!p_{2}!\ldots p_{k}!}\)
Combinations
On many occasions we are not interested in arranging but only in selecting \(r\) objects from given \(n\) objects. A combination is a selection of some or all of a number of different objects where the order of selection is immaterial. The number of selections of \(r\) objects from the given \(n\) objects is denoted by \({ }^n C _r\), and is given by
\(
{ }^n C _r=\frac{n!}{r!(n-r)!}, 0 \leq r \leq n
\)
Remarks
Some important results
Let \(n\) and \(r\) be positive integers such that \(r \leq n\). Then
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The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1 s and exactly three 2 s, is equal to : [JEE Main 2025 (Online) 2nd April Morning Shift]
(b) Step 1: Determine the number of each term
The sequence has a total of 10 terms. There are exactly five 1 s and exactly three 2 s. The remaining terms must be Os.
Total terms, \(\boldsymbol{n} \boldsymbol{=} \mathbf{1 0}\).
Number of \(1 \mathrm{~s}, n_1=5\).
Number of \(2 \mathrm{~s}, n_2=3\).
Number of \(0 \mathrm{~s}, n_0=10-5-3=2\).
Step 2: Apply the multinomial coefficient formula
The number of unique sequences with a given number of repeated elements is given by the multinomial coefficient formula:
\(
\text { Number of sequences }=\frac{n!}{n_{0}!n_{1}!n_{2}!}
\)
Substituting the values:
\(
\frac{10!}{2!5!3!}=\frac{3,628,800}{2 \times 120 \times 6}=\frac{3,628,800}{1440}
\)
Calculating the result:
\(
\frac{3,628,800}{1440}=2520
\)
The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1 s and exactly three 2 s, is equal to 2520.
If all the words with or without meaning made using all the letters of the word “KANPUR” are arranged as in a dictionary, then the word at \(440^{\text {th }}\) position in this arrangement is : [JEE Main 2025 (Online) 29th January Evening Shift]
(b) The letters of “KANPUR” in alphabetical order are A, K, N, P, R, U. The words are ordered in the dictionary as follows:
Words starting with ‘A’ (5! arrangements): 120 words.
Words starting with ‘ K ‘ (5! arrangements): 120 words.
Words starting with ‘N’ (5! arrangements): 120 words.
Total words before ‘ \(\mathrm{P}^{\prime}=120+120+120=360\) words.
The 440th word starts with ‘ P ‘. The remaining 80 positions (440-360) start with ‘ P ‘. Remaining letters: A, K, N, R, U (in alphabetical order).
Words starting with ‘PA’ (4! arrangements): 24 words.
Words starting with ‘PK’ (4! arrangements): 24 words.
Words starting with ‘PN’ (4! arrangements): 24 words.
Total words before ‘PR’ = 360 + 24 + 24 + 24 = 432 words.
The 440th word starts with ‘PR’. The remaining 8 positions (440-432) start with ‘PR’. Remaining letters: A, K, N, U (in alphabetical order).
Words starting with ‘PRA’ (3! arrangements): 6 words.
Words starting with ‘PRK’ (3! arrangements): 6 words.
Total words before ‘PRN’ \(=432+6+6=444\) words.
The 440th word starts with ‘PRK’. The remaining positions ( \(440-438\), since PRA and PRK cover positions 433-438 and 439-444 respectively) are in the ‘PRK’ block. Remaining letters: A, N, U (in alphabetical order). The first word in this block (position 439) is PRKANU. The second word (position 440) is PRKAUN.
The word at the 440th position is PRKAUN.
Alternate: Arranging the letters alphabetically: \(\{\mathrm{A}, \mathrm{K}, \mathrm{N}, \mathrm{P}, \mathrm{R}, \mathrm{U}\}\)
Step 1: Words starting with A = 5!= 120
Step 2: Words starting with \(\mathrm{K}=5!=120\)
Step 3: Words starting with \(\mathrm{N}=5!=120\)
Step 4: Words starting with \(\mathrm{PA}=4!=24\)
Step 5: Words starting with \(\mathrm{PK}=4!=24\)
Step 6: Words starting with \(\mathrm{PN}=4!=24\)
Step 7: Words starting with PRKA \(=3!=6\)
Step 8: PRKAN is the 439th word.
Step 9: PRKAUN is the 440th word.
Let \(P\) be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in \(P\) are formed by using the digits 1,2 and 3 only, then the number of elements in the set \(P\) is : [JEE Main 2025 (Online) 29th January Morning Shift]
(c) (i) number of numbers created using
\(
1111133=\frac{7!}{5!2!} \Rightarrow 21
\)
(ii) number of numbers created using
\(
1111223=\frac{7!}{4!2!} \Rightarrow 105
\)
(iii) number of numbers created using
\(
1112222=\frac{7!}{4!3!} \Rightarrow 35
\)
Total \(=161\)
Let \({ }^n C_{r-1}=28,{ }^n C_r=56\) and \({ }^n C_{r+1}=70\). Let \(A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)\) and \(C\left(3 r-n, r^2-n-1\right)\) be the vertices of a triangle \(A B C\), where \(t\) is a parameter. If \((3 x-1)^2+(3 y)^2=\alpha\), is the locus of the centroid of triangle ABC , then \(\alpha\) equals [JEE Main 2025 (Online) 28th January Morning Shift]
(a) First we determine \(n\) and \(r\) from the given binomial coefficients:
\(
{ }^n C_{r-1}=28, \quad{ }^n C_r=56, \quad{ }^n C_{r+1}=70 .
\)
Step 1: Use ratio relations
\(
\begin{array}{r}
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}=\frac{56}{28}=2 \\
n-r+1=2 r \quad \Rightarrow \quad n=3 r-1
\end{array}
\)
Next,
\(
\begin{gathered}
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}=\frac{70}{56}=\frac{5}{4} \\
4(n-r)=5(r+1)
\end{gathered}
\)
Insert \(n=3 r-1\) :
\(
\begin{gathered}
4(3 r-1-r)=5(r+1) \\
4(2 r-1)=5 r+5 \\
8 r-4=5 r+5 \Rightarrow 3 r=9 \Rightarrow r=3
\end{gathered}
\)
Then \(n=3 r-1=8\).
Step 2: Coordinates of point \(C\)
\(
C\left(3 r-n, r^2-n-1\right)=(1,0)
\)
Step 3: Centroid of triangle \(A B C\)
\(
A=(4 \cos t, 4 \sin t), \quad B=(2 \sin t,-2 \cos t), \quad C=(1,0)
\)
Centroid \(G=(x, y)\) :
\(
x=\frac{4 \cos t+2 \sin t+1}{3}, \quad y=\frac{4 \sin t-2 \cos t}{3}
\)
Step 4: Substitute into locus equation
Given locus:
\(
(3 x-1)^2+(3 y)^2=\alpha
\)
Compute:
\(
3 x-1=4 \cos t+2 \sin t, \quad 3 y=4 \sin t-2 \cos t
\)
Now:
\(
(4 \cos t+2 \sin t)^2+(4 \sin t-2 \cos t)^2
\)
Expand:
\(
=\left(16 \cos ^2 t+16 \cos t \sin t+4 \sin ^2 t\right)+\left(16 \sin ^2 t-16 \cos t \sin t+4 \cos ^2 t\right)
\)
Combine:
\(
=20\left(\cos ^2 t+\sin ^2 t\right)=20
\)
Thus,
\(
\alpha =20.
\)
The number of different 5 digit numbers greater than 50000 that can be formed using the digits \(0,1,2,3,4,5,6,7\), such that the sum of their first and last digits should not be more than 8 , is [JEE Main 2025 (Online) 28th January Morning Shift]
(d) Step 1: Determine the constraints on the first and last digits
The number must be greater than 50,000 , so the first digit ( \(d_1\) ) must be 5,6 , or 7. The available digits are \(\{0,1,2,3,4,5,6,7\}\). The sum of the first and last digits ( \(d_1+d_5\) ) must not be more than 8. Digits can be repeated.
We consider cases based on the first digit:
If \(\boldsymbol{d}_{\mathbf{1}}=\mathbf{5}\) : The condition is \(5+d_5 \leq 8\), so \(d_5 \leq 3\). The possible values for \(d_5\) are \(\{0,1,2,3\}\) (4 options).
If \(\boldsymbol{d}_{\mathbf{1}}=\mathbf{6}\) : The condition is \(\mathbf{6} \boldsymbol{+} \boldsymbol{d}_5 \leq 8\), so \(\boldsymbol{d}_5 \leq 2\). The possible values for \(\boldsymbol{d}_5\) are \(\{0,1,2\}\) (3 options).
If \(\boldsymbol{d}_{\mathbf{1}}=\mathbf{7}\) : The condition is \(7+d_5 \leq 8\), so \(d_5 \leq 1\). The possible values for \(d_5\) are \(\{0,1\}\) (2 options).
Step 2: Determine the possibilities for the middle digits
The middle three digits \(\left(d_2, d_3, d_4\right)\) can be any of the 8 available digits, as repetition is allowed.
Number of choices for \(d_2\) is 8.
Number of choices for \(d_3\) is 8.
Number of choices for \(d_4\) is 8.
The number of possibilities for the middle three digits is \(8 \times 8 \times 8=8^3=512\).
Step 3: Calculate the total number of valid numbers
We sum the number of valid numbers for each case of the first digit:
For \(d_1=5: 4\) (for \(\left.d_5\right) \times 512\left(\right.\) for \(\left.d_2, d_3, d_4\right)=2048\) numbers.
For \(d_1=6: 3\left(\right.\) for \(\left.d_5\right) \times 512\left(\right.\) for \(\left.d_2, d_3, d_4\right)=1536\) numbers.
For \(d_1=7: 2\left(\right.\) for \(\left.d_5\right) \times 512\left(\right.\) for \(\left.d_2, d_3, d_4\right)=1024\) numbers.
The total number of possible 5-digit numbers is the sum of these cases:
\(
\text { Total }=2048+1536+1024=4608
\)
Note that the number 50000 itself is not included because for \(d_1=5\) and \(d_5=0\), we consider numbers like 50000, but the problem asks for numbers greater than 50000 . All the 5-digit numbers generated by this method except 50000 (which has \(d_2=d_3=d_4=0\) ) are greater than 50000 . If 50000 is excluded, the answer is \(4608-1=4607\).
Group \(A\) consists of 7 boys and 3 girls, while group \(B\) consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group \(A\) and the remaining 3 from group \(B\), is equal to : [JEE Main 2025 (Online) 24th January Evening Shift]
(a) Step 1: Define the possible cases
A total of 8 people must be invited: 5 from Group A ( 7 boys, 3 girls) and 3 from Group B ( 6 boys, 5 girls). The final selection must have 4 boys and 4 girls. We consider the possible distributions of boys and girls across the two groups. The number of boys from Group \(\mathrm{A}\left(b_A\right)\) can be 2,3 , or 4.
Case 1: 2 boys and 3 girls from Group A, 2 boys and 1 girl from Group B.
Case 2: 3 boys and 2 girls from Group A, 1 boy and 2 girls from Group B.
Case 3: 4 boys and 1 girl from Group A, 0 boys and 3 girls from Group B.
Step 2: Calculate the combinations for each case
The number of ways for each case is calculated using combinations
\(
\binom{n}{k}=\frac{n!}{k!(n-k)!} .
\)
Case 1: \(\binom{7}{2} \times\binom{ 3}{3} \times\binom{ 6}{2} \times\binom{ 5}{1}=21 \times 1 \times 15 \times 5=1575\)
Case 2: \(\binom{7}{3} \times\binom{ 3}{2} \times\binom{ 6}{1} \times\binom{ 5}{2}=35 \times 3 \times 6 \times 10=6300\)
Case 3: \(\binom{7}{4} \times\binom{ 3}{1} \times\binom{ 6}{0} \times\binom{ 5}{3}=35 \times 3 \times 1 \times 10=1050\)
Step 3: Sum the ways for all cases
The total number of ways is the sum of the ways for each mutually exclusive case.
\(
\text { Total ways }=1575+6300+1050=8925
\)
The number of ways, 4 boys and 4 girls can be invited for a picnic is equal to 8925.
The number of words, which can be formed using all the letters of the word “DAUGHTER”, so that all the vowels never come together, is : [JEE Main 2025 (Online) 23rd January Morning Shift]
(d) Total words \(=8!\)
Total words in which vowels are together \(=6!\times 3!\) words in which all vowels are not together
\(
\begin{aligned}
& =8!-6!\times 3! \\
& =6![56-6] \\
& =720 \times 50 \\
& =36000
\end{aligned}
\)
In a group of 3 girls and 4 boys, there are two boys \(B_1\) and \(B_2\). The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \(B_1\) and \(B_2\) are not adjacent to each other, is: [JEE Main 2025 (Online) 22nd January Evening Shift]
(d) Step 1: Arrange the girl and boy blocks
First, we treat the group of 3 girls (GGG) as one single block and the group of 4 boys (BBBB) as another single block. These two blocks can be arranged in a queue in 2 ! ways. The girls within their block can be arranged in 3 ! ways, and the boys within their block can be arranged in 4 ! ways. The total ways to arrange them with girls together and boys together (without the \(\boldsymbol{B}_1, \boldsymbol{B}_2\) constraint) is:
\(
N_{\text {total }}=2!\times 3!\times 4!=2 \times 6 \times 24=288
\)
Step 2: Calculate adjacent arrangements of \(\boldsymbol{B}_{\mathbf{1}}\) and \(\boldsymbol{B}_{\mathbf{2}}\)
We need to find the arrangements within the boys’ block where the specific boys, \(\boldsymbol{B}_1\) and \(\boldsymbol{B}_2\), are adjacent. We treat the pair ( \(\boldsymbol{B}_1 \boldsymbol{B}_2\) ) as a single unit. We now have 3 items to arrange: the ( \(\boldsymbol{B}_1 \boldsymbol{B}_2\) ) unit, \(\boldsymbol{B}_3\), and \(\boldsymbol{B}_4\). This can be done in 3 ! ways. The boys within the adjacent pair can swap positions (e.g., \(\boldsymbol{B}_1 \boldsymbol{B}_2\) or \(\boldsymbol{B}_2 \boldsymbol{B}_1\) ), which gives 2 ! ways. The number of arrangements with \(\boldsymbol{B}_1\) and \(\boldsymbol{B}_2\) adjacent within the boys’ block is:
\(
N_{\text {adjacent }}=3!\times 2!=6 \times 2=12
\)
Step 3: Calculate non-adjacent arrangements of \(\boldsymbol{B}_{\mathbf{1}}\) and \(\boldsymbol{B}_{\mathbf{2}}\)
The number of ways to arrange the boys’ block such that \(\boldsymbol{B}_1\) and \(\boldsymbol{B}_2\) are not adjacent is the total number of boy arrangements (4!) minus the number of arrangements where they are adjacent ( \(\boldsymbol{N}_{\text {adjacent }}\) ):
\(
N_{\text {non-adjacent }}=4!-N_{\text {adjacent }}=24-12=12
\)
Step 4: Combine the results
Finally, we multiply the number of ways to arrange the girls’ block (3!), the arrangement of the two main blocks (2!), and the number of non-adjacent arrangements of the boys’ block ( \(N_{\text {non-adjacent }}\) ) to get the final answer:
\(
N_{\text {final }}=2!\times 3!\times N_{\text {non-adjacent }}=2 \times 6 \times 12=144
\)
The number of ways in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \(\boldsymbol{B}_1\) and \(\boldsymbol{B}_2\) are not adjacent to each other, is 144.
From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ‘ \(M\) ‘, is : [JEE Main 2025 (Online) 22nd January Morning Shift]
(b) Fix ‘ M ‘: ‘ M ‘ is the 13th letter, so it’s in the middle (3rd) position of the five letters when arranged alphabetically: __ __ M __ __.
Choose Letters Before ‘M’: You need 2 letters from A to L (12 letters). The number of ways to choose is \(\binom{12}{2}\).
\(\binom{12}{2}=\frac{12 \times 11}{2 \times 1}=66\) ways.
Choose Letters After ‘ \(\mathbf{M}\) ‘: You need 2 letters from \(\mathbf{N}\) to \(\mathbf{Z}\) ( 13 letters). The number of ways to choose is \(\binom{13}{2}\).
\(\binom{13}{2}=\frac{13 \times 12}{2 \times 1}=78\) ways.
Total Ways: Multiply the choices for letters before and after ‘M’.
Total ways \(=66 \times 78=5148\).
Let m and \(\mathrm{n},(\mathrm{m}<\mathrm{n})\), be two 2-digit numbers. Then the total numbers of pairs \((\mathrm{m}, \mathrm{n})\), such that \(\operatorname{gcd}(m, n)=6\), is ____. [JEE Main 2025 (Online) 4th April Evening Shift]
(a) Step 1: Understand the given conditions
We are looking for pairs of 2-digit integers ( \(m, n\) ) that satisfy:
\(10 \leq m, n \leq 99\)
\(m<n\)
\(\operatorname{gcd}(m, n)=6\)
Step 2: Formulate the numbers in terms of their GCD
Since \(\operatorname{gcd}(m, n)=6\), we can express \(m\) and \(n\) as:
\(m=6 a\)
\(n=6 b\)
where \(a\) and \(b\) are positive integers with \(\operatorname{gcd}(a, b)=1\).
Step 3: Determine the possible range for \(\boldsymbol{a}\) and \(\boldsymbol{b}\)
Using the range for 2-digit numbers, we find the constraints for \(a\) and \(b\) :
\(10 \leq 6 a \leq 99 \Longrightarrow \frac{10}{6} \leq a \leq \frac{99}{6} \Longrightarrow 1.67 \leq a \leq 16.5\). Thus, \(a \in\{2,3, \ldots, 16\}\).
\(10 \leq 6 b \leq 99 \Longrightarrow \frac{10}{6} \leq b \leq \frac{99}{6} \Longrightarrow 1.67 \leq b \leq 16.5\). Thus, \(b \in\{2,3, \ldots, 16\}\)
The condition \(m<n\) implies \(6 a<6 b\), so \(a<b\). Therefore, we need to count the number of pairs \((a, b)\) from the set \(S=\{2,3, \ldots, 16\}\) such that \(a<b\) and \(\operatorname{gcd}(a, b)=1\).
Step 4: Count the valid coprime pairs
By enumerating all possible coprime pairs ( \(a, b\) ) within the determined range (using a computational check), the total count is found to be 64.
The total numbers of pairs \((m, n)\) such that \(\operatorname{gcd}(m, n)=6\) is 64.
Explanation: Step 1: Define the conditions for the numbers \(\boldsymbol{m}\) and \(\boldsymbol{n}\)
The numbers \(m\) and \(n\) are 2-digit numbers, which means they must satisfy \(10 \leq m, n \leq 99\). We are also given \(m<n\) and \(\operatorname{gcd}(m, n)=6\).
Step 2: Express \(\boldsymbol{m}\) and \(\boldsymbol{n}\) in terms of their greatest common divisor
Since \(\operatorname{gcd}(m, n)=6\), we can write \(m=6 a\) and \(n=6 b\), where \(a\) and \(b\) are integers such that \(\operatorname{gcd}(a, b)=1\) and \(a<b\).
Step 3: Determine the possible ranges for \(\boldsymbol{a}\) and \(\boldsymbol{b}\)
Using the range for \(m\) and \(n\) :
\(10 \leq 6 a \leq 99 \Longrightarrow 1.67 \leq a \leq 16.5\). The possible integer values for \(a\) are \(\{2,3, \ldots, 16\}\).
\(10 \leq 6 b \leq 99 \Longrightarrow 1.67 \leq b \leq 16.5\). The possible integer values for \(b\) are \(\{2,3, \ldots, 16\}\).
Both \(a\) and \(b\) can be any integer from 2 to 16 , but we must also satisfy \(a<b\) and \(\operatorname{gcd}(a, b)=1\).
Step 4: Count the valid pairs ( \(a, b\) )
We need to count the number of pairs ( \(a, b\) ) such that \(2 \leq a<b \leq 16\) and their greatest common divisor is 1. By systematically listing or using computational methods, the number of such coprime pairs is 64. The pairs are the following:
\((2,3),(2,5),(2,7),(2,9),(2,11),(2,13),(2,15)\)
\((3,4),(3,5),(3,7),(3,8),(3,10),(3,11),(3,13),(3,14),(3,16)\)
\((4,5),(4,7),(4,9),(4,11),(4,13),(4,15)\)
\((5,6),(5,7),(5,8),(5,9),(5,11),(5,12),(5,13),(5,14),(5,16)\)
\((6,7),(6,11),(6,13)\)
\((7,8),(7,9),(7,10),(7,11),(7,12),(7,13),(7,15),(7,16)\)
\((8,9),(8,11),(8,13),(8,15)\)
\((9,10),(9,11),(9,13),(9,14),(9,16)\)
\((10,11),(10,13)\)
\((11, 12), (11, 13), (11, 14), (11, 15), (11, 16)\)
\((12,13)\)
\((13,14),(13,15),(13,16)\)
\((14,15)\)
\((15,16)\)
The total numbers of pairs \((m, n)\), such that \(\operatorname{gcd}(m, n)=6\), is 64.
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number \(n\) be denoted by \(\mathrm{W}_{\mathrm{n}}\). Let the probability \(\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)\) of choosing the word \(\mathrm{W}_{\mathrm{n}}\) satisfy \(\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}>1\).
If \(\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}\), then \(\alpha+\beta\) is equal to : ____. [JEE Main 2025 (Online) 3rd April Morning Shift]
(d) The total number of permutations (words) is \(5!=120\). Let the serial number of CDBEA be \(n\).
The words are ordered alphabetically, meaning we count permutations starting with letters alphabetically before ‘C’.
Number of words starting with \(\mathrm{A}=4!=24\).
Number of words starting with \(\mathrm{B}=4!=24\).
Next are words starting with c. We look at the second letter, counting those starting with CA and CB.
Number of words starting with \(\mathrm{CA}=3!=6\).
Number of words starting with \(\mathrm{CB}=3!=6\).
Next are words starting with cc, which are not possible. Next are words starting with CD . We look at the third letter, counting those starting with CDA and CDB .
Number of words starting with \(\mathrm{CDA}=2!=2\).
Next are words starting with CDB . The remaining letters are A, E. The alphabetical order is CDBAE, then CDBEA.
CDBAE is the first word starting with CDB .
CDBEA is the second word starting with CDB .
The serial number \(n\) of CDBEA is \(24+24+6+6+2+2=64\).
The probabilities form a geometric series where \(P\left(W_n\right)=2^{n-1} P\left(W_1\right)\).
The sum of all probabilities is 1:
\(
\begin{aligned}
& \sum_{i=1}^{120} P\left(W_i\right)=P\left(W_1\right)\left(1+2+2^2+\ldots+2^{119}\right)=P\left(W_1\right)\left(2^{120}-1\right)=1 . \\
& P\left(W_1\right)=\frac{1}{2^{120}-1} . \\
& P(\mathrm{CDBEA})=P\left(W_{64}\right)=2^{63} P\left(W_1\right)=\frac{2^{63}}{2^{120}-1} .
\end{aligned}
\)
Comparing this to the given form \(\frac{2^\alpha}{2^\beta-1}\), we have \(\alpha=63\) and \(\beta=120\).
\(
\alpha+\beta=63+120=183 .
\)
If the number of seven-digit numbers, such that the sum of their digits is even, is \(m \cdot n \cdot 10^n ; m, n \in\{1,2,3, \ldots, 9\}\), then \(m+n\) is equal to ____. [JEE Main 2025 (Online) 3rd April Morning Shift]
(b) Step 1: Calculate the number of seven-digit numbers with an even digit sum
A seven-digit number has digits \(d_1 d_2 d_3 d_4 d_5 d_6 d_7\), where \(d_1 \in\{1,2, \ldots, 9\}\) and \(d_i \in\{0,1, \ldots, 9\}\) for \(i \in\{2, \ldots, 7\}\). To determine the count of numbers with an even digit sum, we note that for any combination of the first six digits ( \(9 \times 10^5\) possibilities), there are exactly 5 choices for the seventh digit that make the total sum even. The total number of such seven-digit numbers is calculated as:
\(
N=9 \times 10^5 \times 5=45 \times 10^5
\)
Step 2: Determine the values of \(m\) and \(n\)
The problem states that the number of these seven-digit numbers is \(m \cdot n \cdot 10^n\), where \(m, n \in\{1,2, \ldots, 9\}\). We equate our calculated number to this expression:
\(
45 \times 10^5=m \cdot n \cdot 10^n
\)
By comparing the terms, we can infer that \(n=5\). This leaves \(m \cdot n=45\), so \(m \cdot 5=45\), which gives \(m=9\). Both \(m=9\) and \(n=5\) satisfy the condition \(m, n \in\{1,2, \ldots, 9\}\). The value of \(m+n\) is then calculated:
\(
m+n=9+5=14
\)
The value of \(m+n\) is 14.
Explanation: Step 1: Calculate the total number of seven-digit numbers
The first digit can be any number from 1 to 9 ( 9 options), and the remaining six digits can be any number from 0 to 9 ( 10 options each).
The total number of seven-digit numbers is \(9 \times 10^6\).
Step 2: Determine the count of numbers with an even digit sum
For any multi-digit number, exactly half have an even digit sum and half have an odd digit sum due to the principle of parity.
The number of seven-digit numbers with an even digit sum is:
\(
\frac{9 \times 10^6}{2}=4,500,000
\)
Step 3: Find the values of \(m\) and \(n\)
We are given that the number is \(m \cdot n \cdot 10^n\).
\(
4,500,000=4.5 \times 10^6
\)
We can rewrite this in the form \(m \cdot n \cdot 10^n\).
One way is \(9 \cdot 5 \cdot 10^5\), where \(m=9\) and \(n=5\). Both \(m, n \in\{1,2, \ldots, 9\}\).
Step 4: Calculate \(\boldsymbol{m} \boldsymbol{+} \boldsymbol{n}\)
Using the values \(m=9\) and \(n=5\) :
\(
m+n=9+5=14
\)
The number of 6 -letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is ____. [JEE Main 2025 (Online) 29th January Morning Shift]
(d)
(6): One letter appears 6 times. \(\binom{5}{1} \times \frac{6!}{6!}=5 \times 1=5\) ways.
(4, 2): One letter 4 times, another 2 times. \(\binom{5}{1} \times\binom{ 4}{1} \times \frac{6!}{4!2!}=5 \times 4 \times 15=300\) ways.
(3,3) : Two letters appear 3 times each. \(\binom{5}{2} \times \frac{6!}{3!3!}=10 \times 20=200\) ways.
(2,2,2) : Three letters appear 2 times each. \(\binom{5}{3} \times \frac{6!}{2!2!2!}=10 \times 90=900\)
ways.
\(
\text { Total }=5+300+200+900=1405 .
\)
Explanation: 1. Letters available: M, A, T, H, S (5 distinct letters).
2. Word length: 6 letters.
3. Condition: Any letter used must appear at least twice.
This leads to three possible compositions for the 6-letter word:
Case 1: One letter used (repeated 6 times)
Choose 1 letter from 5: \(\binom{5}{1}=5\) ways.
Arrange: \(\frac{6!}{6!}=1\) way (e.g., MMMMMM).
Total: \(5 \times 1=5\) words.
Case 2: Two distinct letters used (patterns like 4, 2 or 3, 3)
Choose 2 letters from 5: \(\binom{5}{2}=10\) ways.
For each pair (say X, Y):
Pattern \((4,2): \frac{6!}{4!2!} \times 2=15 \times 2=30\) (4 X’s, 2 Y’s OR 4 Y’s, 2 X’s).
Pattern \((3,3): \frac{6!}{3!3!}=20\) ways ( 3 X’s, 3 Y’s).
Total for this case: \(10 \times(30+20)=10 \times 50=500\) words.
Case 3: Three distinct letters used (each appearing twice – pattern 2, 2, 2)
Choose 3 letters from 5: \(\binom{5}{3}=10\) ways.
Arrange: \(\frac{6!}{2!2!2!}=\frac{720}{8}=90\) ways (e.g., MMTTAA).
Total for this case: \(10 \times 90=900\) words.
Grand Total: \(5+500+900=1405\) words.
The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _____. [JEE Main 2025 (Online) 28th January Evening Shift]
(b) Step 1: Understand the Problem
We are given a three-digit number \(x=\overline{x y z}\), where \(x, y, z\) are digits. The condition is:
\(
x+y+z=15
\)
Additionally, since \(x\) is the hundreds digit, \(x\) must satisfy \(2 \leq x \leq 9\).
Step 2: Identify Possible Combinations for Each \(x\)
For \(x=2, y+z=13 \rightarrow 6\) combinations: \((4,9),(5,8),(6,7),(7,6),(8,5),(9,4)\)
For \(x=3, y+z=12 \rightarrow 7\) combinations: \((3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)\)
For \(x=4, y+z=11 \rightarrow 8\) combinations: \((2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)\)
For \(x=5, y+z=10 \rightarrow 9\) combinations: \((1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8\), 2), \((9,1)\)
For \(x=6, y+z=9 \rightarrow 10\) combinations: \((0,9),(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7\), 2), (8, 1), (9, 0)
For \(x=7, y+z=8 \rightarrow 9\) combinations: \((0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1)\), \((8,0)\)
For \(x=8, y+z=7 \rightarrow 8\) combinations: \((0,7),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(7,0)\)
For \(x=9, y+z=6 \rightarrow 7\) combinations: \((0,6),(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)\)
Step 3: Calculate the Total Number of Possible Values
The total number of valid combinations is:
\(
6+7+8+9+10+9+8+7=64
\)
Number of functions \(f:\{1,2, \ldots, 100\} \rightarrow\{0,1\}\), that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to ____. [JEE Main 2025 (Online) 24th January Evening Shift]
(c) Step 1: Choose the number \(\leq 98\) that gets 1
Numbers 1 through \(98 \rightarrow\) choose exactly one to assign 1.
\(
\text { Choices }=98
\)
Step 2: Assign values to remaining numbers
Remaining numbers \(\leq 98\) : 97 numbers \(\rightarrow\) must be 0
Numbers 99 and \(100 \rightarrow\) can be 0 or 1 independently \(\rightarrow 2\) choices each \(\rightarrow 2 \times 2=4\)
Step 3: Total number of functions
\(
98 \times 4=392
\)
The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by 4 and 9 , is ____. [JEE Main 2025 (Online) 24th January Morning Shift]
(a) No, of 3 digits \(=999-99=900\)
No. of 3 digit numbers divisible by 2 & 3 i.e. by 6
\(
\frac{900}{6}=150
\)
No. of 3 digit numbers divisible by 4 & 9 i.e. by 36
\(
\frac{900}{36}=25
\)
∴ No of 3 digit numbers divisible by \(2 \& 3\) but not by \(4 \& 9\)
\(
150-25=125
\)
The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is ____. [JEE Main 2025 (Online) 23rd January Evening Shift]
(c) Case 1: All boys sit together.
Treat the 5 boys as a single unit or block.
This results in 1 “boy-block” and 4 girls, totaling 5 units to arrange, which can be done in \(5!\) ways.
Within the boy-block, the 5 boys can be arranged among themselves in 5 ! ways.
Total arrangements for Case 1: \(5!\times 5!=120 \times 120=14,400\) ways.
Case 2: No two boys sit together.
Arrange the \(\mathbf{4}\) girls first in 4 ! ways.
This creates 5 possible spaces (gaps) where the boys can sit to ensure no two are adjacent: _G_G_G_G_ .
Place the 5 boys into these 5 gaps in \(P(5,5)=5!\) ways.
Total arrangements for Case 2: \(4!\times 5!=24 \times 120=2,880\) ways.
Total Ways:
Sum the ways from both cases: \(14,400+2,880=17,280\) ways.
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